Eigenvectors of Linear Transformations
Markus Stroppel
The following examples attempt to show linear transformations at work, represented as multiplication by matrices A, B, C and D, respectively. We will indicate a vector v (in black)
together with its image (in yellow). Moreover, a triangle (pale green) is shown together with
its image under the transformation. Note that we follow Poole (Linear algebra, Boston, 2011)
in calling every linear map a ‘transformation’, irrespective of its invertibility. In the examples
below, we will give the effect of our geometric transformation by applying matrices to columns.
These columns describe the position vectors with respect to a coordinate system that will be
indicated if necessary.
The one-dimensional space span(v) is indicated by a very thick gray line in the drawings.
In the interactive HTML-version1 you may use your mouse to move v (drag its tip); the image
of v under the respective transformation will move automatically.
1 Reflection, followed by scaling
Our black vector v is mapped to the yellow vector Av by reflecting in the thin red line and
then scaling by the factor 1/2.
If we introduce cartesian coordinates with axes as indicated by the red and blue line, this
transformation is affected by multiplication with the matrix (product)
"
A=
1
2
0
0 − 12
#
"
=
1
2
0
0
1
2
#"
1 0
0 −1
#
.
Turn the black vector v such that it lies on the red line. Then the image Av will also lie on
that line. It will point in the same direction as v but has only half of its length:
We have thus found an eigenvector for
the eigenvalue 12 .
This relation is preserved if you scale
the vector v but keep it on the red
line.
v
If you move v such that it lies on the
blue line then the image Av will also
lie on that line. Again, the image will
have half of the length of v but it will
point in the opposite direction:
Av
We have thus found an eigenvector for
the eigenvalue − 21 .
1
You find that version using the address
http://www.mathematik.uni-stuttgart.de/studium/infomat/HM-Stroppel-Material/Eigenvectors/.
1
It is just a coincidence (in fact, a consequence of the fact that A is symmetric) that the
eigenspaces are orthogonal to each other).
The interactive version also allows to move the right line.
2 Projection (that is, a non-invertible transformation)
In the following example we map v to Bv obtained by projection along the direction of the
green line, onto the red line2 . Find the directions of the eigenvectors!
Vectors pointing along the red line
will not be changed at all by this
projection.
v
These are eigenvectors for the
eigenvalue 1.
Bv
Vectors pointing along the green
line (viz, the direction of our projection) are eigenvector for the
eigenvalue 0:
These vectors are mapped to the
zero vector.
Matrix examples for such projections include the following.
"
• B=
0 0
0 1
#
(if the green and red lines form the first and second axis of a cartesian coordinate system,
respectively).

• B=
3
5
4
5


h

3
5
4
5
i
=
9
25
12
25
12
25
16
25



(if the red line is spanned by the vector u = 
3
5
4
5

,
and the green line is perpendicular
to the red one).
"
• B=
5 −2
10 −4
#
"
(here the red line is spanned by
1
2
#
"
and the green one by
perpendicular).
2
You may change these directions in the interactive version.
2
#
2
; these lines are not
5
There are many more
of 2 × 2 matrices with eigenvalues 0 and 1; in fact, these are
" examples
#
a b
just those matrices
where a + d = 1 and ad − bc = 0.
c d
In other words, the trace and determinant of B have to satisfy tr(B) = 1 and det(B) = 0.
This may also be phrased as: The characterstic polynomial det(B − λI) of B equals λ2 − λ.
3 Shearing (algebraic and geometric multiplicities differ)
In the following example, we do not find (in fact, we do not have) eigenvectors in different
directions.
Our shearing fixes each point on the red line3 ; every other point is then moved on the parallel
to the red line through that point. Find the directions of the eigenvectors!
Vectors pointing along the red
line will not be changed by the
shear:
Cv
These are eigenvectors for the
eigenvalue 1.
v
As every other vector changes his
direction under the shear, there
are no other eigenvectors.
Among the matrices describing such shears, we have:
"
• C=
"
• C=
1 0
3 1
#
−1 2
−2 3
(here the red line is the second coordinate axis).
#
"
(the red line is spanned by
#
1
.)
1
The characteristic property of these matrices describing shears is the fact that they have the
eigenvalue 1 with algebraic multiplicity 2 while the geometric multiplicity is only 1.
There are many" more #examples of 2 × 2 matrices with this property; in fact, these are just
a b
those matrices
that are not diagonal (i.e., at least one of the entries b, c is different
c d
from zero) with a + d = 2 and ad − bc = 1.
In other words, the trace and determinant of C have to satisfy tr(C) = 2 and det(C) = 1.
This may also be phrased as follows:
The characterstic polynomial det(C − λI) of C equals λ2 − 2λ + 1 = (λ − 1)2 .
3
You may change this line and the amount of shearing in the interactive version.
3
4 Rotation (no real eigenvalues, but imaginary ones)
Finally, we discuss an example of a transformation that can be described by a matrix with
real entries but having no real eigenvalue at all:
We rotate v.
(The interactive version allows to
change the angle of rotation.)
Dv
Under the rotation indicated in
the drawing, every non-zero vector will change its direction.
Therefore, there are no eigenvectors to be seen.
v
However — can you think of other
angles of rotation where we do see
eigenvectors?
The drawing indicates a counterclockwise rotation by 60◦ , or
described by the matrix
√ 

− 3
1
D=
2
√
3
2
π
3
radians. That rotation can be
2
1
2

The √characteristic polynomial is det(D − λI) = λ2 − λ + 1; the eigenvalues are
1
− 23 i.
2
1
2
√
+
3
i
2
and
There are many more examples
"
#of 2 × 2 matrices without real eigenvalues; in fact, these are
a b
just those matrices D =
where the characteristic equation
c d
0 = det(D − λI) = λ2 − (a + d)λ + (ad − bc)
has no solution in the field of real numbers.
This happens precisely if the discriminant (a + d)2 − 4(ad − bc) is negative; i.e., if
(a − d)2 < −4bc .
(For the rotation in the drawing, we have a = d and bc < 0, and the condition is satisfied.)
4